Integrand size = 28, antiderivative size = 133 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 \left (b^2-4 a c\right ) d (b d+2 c d x)^{26/3}}+\frac {9 \left (a+b x+c x^2\right )^{7/3}}{65 \left (b^2-4 a c\right )^2 d^3 (b d+2 c d x)^{20/3}}+\frac {27 \left (a+b x+c x^2\right )^{7/3}}{455 \left (b^2-4 a c\right )^3 d^5 (b d+2 c d x)^{14/3}} \] Output:
3/13*(c*x^2+b*x+a)^(7/3)/(-4*a*c+b^2)/d/(2*c*d*x+b*d)^(26/3)+9/65*(c*x^2+b *x+a)^(7/3)/(-4*a*c+b^2)^2/d^3/(2*c*d*x+b*d)^(20/3)+27/455*(c*x^2+b*x+a)^( 7/3)/(-4*a*c+b^2)^3/d^5/(2*c*d*x+b*d)^(14/3)
Time = 5.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 (a+x (b+c x))^{7/3} \left (65 b^4+156 b^3 c x+48 b c^2 x \left (-7 a+6 c x^2\right )+4 b^2 c \left (-91 a+75 c x^2\right )+16 c^2 \left (35 a^2-21 a c x^2+9 c^2 x^4\right )\right )}{455 \left (b^2-4 a c\right )^3 d^9 (b+2 c x)^8 (d (b+2 c x))^{2/3}} \] Input:
Integrate[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3),x]
Output:
(3*(a + x*(b + c*x))^(7/3)*(65*b^4 + 156*b^3*c*x + 48*b*c^2*x*(-7*a + 6*c* x^2) + 4*b^2*c*(-91*a + 75*c*x^2) + 16*c^2*(35*a^2 - 21*a*c*x^2 + 9*c^2*x^ 4)))/(455*(b^2 - 4*a*c)^3*d^9*(b + 2*c*x)^8*(d*(b + 2*c*x))^(2/3))
Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1117, 1117, 1106}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {6 \int \frac {\left (c x^2+b x+a\right )^{4/3}}{(b d+2 c x d)^{23/3}}dx}{13 d^2 \left (b^2-4 a c\right )}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}}\) |
\(\Big \downarrow \) 1117 |
\(\displaystyle \frac {6 \left (\frac {3 \int \frac {\left (c x^2+b x+a\right )^{4/3}}{(b d+2 c x d)^{17/3}}dx}{10 d^2 \left (b^2-4 a c\right )}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}}\right )}{13 d^2 \left (b^2-4 a c\right )}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}}\) |
\(\Big \downarrow \) 1106 |
\(\displaystyle \frac {6 \left (\frac {9 \left (a+b x+c x^2\right )^{7/3}}{70 d^3 \left (b^2-4 a c\right )^2 (b d+2 c d x)^{14/3}}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{10 d \left (b^2-4 a c\right ) (b d+2 c d x)^{20/3}}\right )}{13 d^2 \left (b^2-4 a c\right )}+\frac {3 \left (a+b x+c x^2\right )^{7/3}}{13 d \left (b^2-4 a c\right ) (b d+2 c d x)^{26/3}}\) |
Input:
Int[(a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3),x]
Output:
(3*(a + b*x + c*x^2)^(7/3))/(13*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(26/3)) + (6*((3*(a + b*x + c*x^2)^(7/3))/(10*(b^2 - 4*a*c)*d*(b*d + 2*c*d*x)^(20/3) ) + (9*(a + b*x + c*x^2)^(7/3))/(70*(b^2 - 4*a*c)^2*d^3*(b*d + 2*c*d*x)^(1 4/3))))/(13*(b^2 - 4*a*c)*d^2)
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* c))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & & (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) /2])
Time = 1.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05
method | result | size |
gosper | \(-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (2 c x +b \right ) \left (144 c^{4} x^{4}+288 b \,c^{3} x^{3}-336 a \,c^{3} x^{2}+300 b^{2} c^{2} x^{2}-336 a b \,c^{2} x +156 b^{3} c x +560 a^{2} c^{2}-364 c a \,b^{2}+65 b^{4}\right )}{455 \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (2 c d x +b d \right )^{\frac {29}{3}}}\) | \(139\) |
orering | \(-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {7}{3}} \left (2 c x +b \right ) \left (144 c^{4} x^{4}+288 b \,c^{3} x^{3}-336 a \,c^{3} x^{2}+300 b^{2} c^{2} x^{2}-336 a b \,c^{2} x +156 b^{3} c x +560 a^{2} c^{2}-364 c a \,b^{2}+65 b^{4}\right )}{455 \left (64 a^{3} c^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \left (2 c d x +b d \right )^{\frac {29}{3}}}\) | \(139\) |
Input:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x,method=_RETURNVERBOSE)
Output:
-3/455*(c*x^2+b*x+a)^(7/3)*(2*c*x+b)*(144*c^4*x^4+288*b*c^3*x^3-336*a*c^3* x^2+300*b^2*c^2*x^2-336*a*b*c^2*x+156*b^3*c*x+560*a^2*c^2-364*a*b^2*c+65*b ^4)/(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/(2*c*d*x+b*d)^(29/3)
Leaf count of result is larger than twice the leaf count of optimal. 687 vs. \(2 (115) = 230\).
Time = 0.32 (sec) , antiderivative size = 687, normalized size of antiderivative = 5.17 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\frac {3 \, {\left (144 \, c^{6} x^{8} + 576 \, b c^{5} x^{7} + 12 \, {\left (85 \, b^{2} c^{4} - 4 \, a c^{5}\right )} x^{6} + 65 \, a^{2} b^{4} - 364 \, a^{3} b^{2} c + 560 \, a^{4} c^{2} + 36 \, {\left (29 \, b^{3} c^{3} - 4 \, a b c^{4}\right )} x^{5} + {\left (677 \, b^{4} c^{2} - 196 \, a b^{2} c^{3} + 32 \, a^{2} c^{4}\right )} x^{4} + 2 \, {\left (143 \, b^{5} c - 76 \, a b^{3} c^{2} + 32 \, a^{2} b c^{3}\right )} x^{3} + {\left (65 \, b^{6} + 78 \, a b^{4} c - 540 \, a^{2} b^{2} c^{2} + 784 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left (65 \, a b^{5} - 286 \, a^{2} b^{3} c + 392 \, a^{3} b c^{2}\right )} x\right )} {\left (2 \, c d x + b d\right )}^{\frac {1}{3}} {\left (c x^{2} + b x + a\right )}^{\frac {1}{3}}}{455 \, {\left (512 \, {\left (b^{6} c^{9} - 12 \, a b^{4} c^{10} + 48 \, a^{2} b^{2} c^{11} - 64 \, a^{3} c^{12}\right )} d^{10} x^{9} + 2304 \, {\left (b^{7} c^{8} - 12 \, a b^{5} c^{9} + 48 \, a^{2} b^{3} c^{10} - 64 \, a^{3} b c^{11}\right )} d^{10} x^{8} + 4608 \, {\left (b^{8} c^{7} - 12 \, a b^{6} c^{8} + 48 \, a^{2} b^{4} c^{9} - 64 \, a^{3} b^{2} c^{10}\right )} d^{10} x^{7} + 5376 \, {\left (b^{9} c^{6} - 12 \, a b^{7} c^{7} + 48 \, a^{2} b^{5} c^{8} - 64 \, a^{3} b^{3} c^{9}\right )} d^{10} x^{6} + 4032 \, {\left (b^{10} c^{5} - 12 \, a b^{8} c^{6} + 48 \, a^{2} b^{6} c^{7} - 64 \, a^{3} b^{4} c^{8}\right )} d^{10} x^{5} + 2016 \, {\left (b^{11} c^{4} - 12 \, a b^{9} c^{5} + 48 \, a^{2} b^{7} c^{6} - 64 \, a^{3} b^{5} c^{7}\right )} d^{10} x^{4} + 672 \, {\left (b^{12} c^{3} - 12 \, a b^{10} c^{4} + 48 \, a^{2} b^{8} c^{5} - 64 \, a^{3} b^{6} c^{6}\right )} d^{10} x^{3} + 144 \, {\left (b^{13} c^{2} - 12 \, a b^{11} c^{3} + 48 \, a^{2} b^{9} c^{4} - 64 \, a^{3} b^{7} c^{5}\right )} d^{10} x^{2} + 18 \, {\left (b^{14} c - 12 \, a b^{12} c^{2} + 48 \, a^{2} b^{10} c^{3} - 64 \, a^{3} b^{8} c^{4}\right )} d^{10} x + {\left (b^{15} - 12 \, a b^{13} c + 48 \, a^{2} b^{11} c^{2} - 64 \, a^{3} b^{9} c^{3}\right )} d^{10}\right )}} \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x, algorithm="fricas")
Output:
3/455*(144*c^6*x^8 + 576*b*c^5*x^7 + 12*(85*b^2*c^4 - 4*a*c^5)*x^6 + 65*a^ 2*b^4 - 364*a^3*b^2*c + 560*a^4*c^2 + 36*(29*b^3*c^3 - 4*a*b*c^4)*x^5 + (6 77*b^4*c^2 - 196*a*b^2*c^3 + 32*a^2*c^4)*x^4 + 2*(143*b^5*c - 76*a*b^3*c^2 + 32*a^2*b*c^3)*x^3 + (65*b^6 + 78*a*b^4*c - 540*a^2*b^2*c^2 + 784*a^3*c^ 3)*x^2 + 2*(65*a*b^5 - 286*a^2*b^3*c + 392*a^3*b*c^2)*x)*(2*c*d*x + b*d)^( 1/3)*(c*x^2 + b*x + a)^(1/3)/(512*(b^6*c^9 - 12*a*b^4*c^10 + 48*a^2*b^2*c^ 11 - 64*a^3*c^12)*d^10*x^9 + 2304*(b^7*c^8 - 12*a*b^5*c^9 + 48*a^2*b^3*c^1 0 - 64*a^3*b*c^11)*d^10*x^8 + 4608*(b^8*c^7 - 12*a*b^6*c^8 + 48*a^2*b^4*c^ 9 - 64*a^3*b^2*c^10)*d^10*x^7 + 5376*(b^9*c^6 - 12*a*b^7*c^7 + 48*a^2*b^5* c^8 - 64*a^3*b^3*c^9)*d^10*x^6 + 4032*(b^10*c^5 - 12*a*b^8*c^6 + 48*a^2*b^ 6*c^7 - 64*a^3*b^4*c^8)*d^10*x^5 + 2016*(b^11*c^4 - 12*a*b^9*c^5 + 48*a^2* b^7*c^6 - 64*a^3*b^5*c^7)*d^10*x^4 + 672*(b^12*c^3 - 12*a*b^10*c^4 + 48*a^ 2*b^8*c^5 - 64*a^3*b^6*c^6)*d^10*x^3 + 144*(b^13*c^2 - 12*a*b^11*c^3 + 48* a^2*b^9*c^4 - 64*a^3*b^7*c^5)*d^10*x^2 + 18*(b^14*c - 12*a*b^12*c^2 + 48*a ^2*b^10*c^3 - 64*a^3*b^8*c^4)*d^10*x + (b^15 - 12*a*b^13*c + 48*a^2*b^11*c ^2 - 64*a^3*b^9*c^3)*d^10)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\text {Timed out} \] Input:
integrate((c*x**2+b*x+a)**(4/3)/(2*c*d*x+b*d)**(29/3),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (115) = 230\).
Time = 0.26 (sec) , antiderivative size = 783, normalized size of antiderivative = 5.89 \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx =\text {Too large to display} \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x, algorithm="maxima")
Output:
3/455*(144*c^6*d^(1/3)*x^8 + 576*b*c^5*d^(1/3)*x^7 + 12*(85*b^2*c^4*d^(1/3 ) - 4*a*c^5*d^(1/3))*x^6 + 65*a^2*b^4*d^(1/3) - 364*a^3*b^2*c*d^(1/3) + 56 0*a^4*c^2*d^(1/3) + 36*(29*b^3*c^3*d^(1/3) - 4*a*b*c^4*d^(1/3))*x^5 + (677 *b^4*c^2*d^(1/3) - 196*a*b^2*c^3*d^(1/3) + 32*a^2*c^4*d^(1/3))*x^4 + 2*(14 3*b^5*c*d^(1/3) - 76*a*b^3*c^2*d^(1/3) + 32*a^2*b*c^3*d^(1/3))*x^3 + (65*b ^6*d^(1/3) + 78*a*b^4*c*d^(1/3) - 540*a^2*b^2*c^2*d^(1/3) + 784*a^3*c^3*d^ (1/3))*x^2 + 2*(65*a*b^5*d^(1/3) - 286*a^2*b^3*c*d^(1/3) + 392*a^3*b*c^2*d ^(1/3))*x)*(c*x^2 + b*x + a)^(1/3)/((b^14*d^10 - 12*a*b^12*c*d^10 + 48*a^2 *b^10*c^2*d^10 - 64*a^3*b^8*c^3*d^10 + 256*(b^6*c^8*d^10 - 12*a*b^4*c^9*d^ 10 + 48*a^2*b^2*c^10*d^10 - 64*a^3*c^11*d^10)*x^8 + 1024*(b^7*c^7*d^10 - 1 2*a*b^5*c^8*d^10 + 48*a^2*b^3*c^9*d^10 - 64*a^3*b*c^10*d^10)*x^7 + 1792*(b ^8*c^6*d^10 - 12*a*b^6*c^7*d^10 + 48*a^2*b^4*c^8*d^10 - 64*a^3*b^2*c^9*d^1 0)*x^6 + 1792*(b^9*c^5*d^10 - 12*a*b^7*c^6*d^10 + 48*a^2*b^5*c^7*d^10 - 64 *a^3*b^3*c^8*d^10)*x^5 + 1120*(b^10*c^4*d^10 - 12*a*b^8*c^5*d^10 + 48*a^2* b^6*c^6*d^10 - 64*a^3*b^4*c^7*d^10)*x^4 + 448*(b^11*c^3*d^10 - 12*a*b^9*c^ 4*d^10 + 48*a^2*b^7*c^5*d^10 - 64*a^3*b^5*c^6*d^10)*x^3 + 112*(b^12*c^2*d^ 10 - 12*a*b^10*c^3*d^10 + 48*a^2*b^8*c^4*d^10 - 64*a^3*b^6*c^5*d^10)*x^2 + 16*(b^13*c*d^10 - 12*a*b^11*c^2*d^10 + 48*a^2*b^9*c^3*d^10 - 64*a^3*b^7*c ^4*d^10)*x)*(2*c*x + b)^(2/3))
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{\frac {4}{3}}}{{\left (2 \, c d x + b d\right )}^{\frac {29}{3}}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^(4/3)/(2*c*d*x + b*d)^(29/3), x)
Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^{4/3}}{{\left (b\,d+2\,c\,d\,x\right )}^{29/3}} \,d x \] Input:
int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3),x)
Output:
int((a + b*x + c*x^2)^(4/3)/(b*d + 2*c*d*x)^(29/3), x)
\[ \int \frac {\left (a+b x+c x^2\right )^{4/3}}{(b d+2 c d x)^{29/3}} \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{\frac {4}{3}}}{\left (2 c d x +b d \right )^{\frac {29}{3}}}d x \] Input:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x)
Output:
int((c*x^2+b*x+a)^(4/3)/(2*c*d*x+b*d)^(29/3),x)